Exercises — Doppler effect — all cases - source moving, observer moving, both

Upar. Tum observer ho, aur source ruka hua hai, isliye hawa mein floating wavelength bilkul untouched hai — crest-arrival speed badlti hai. Crests mein ghuste hue, tum unhe zyada tezi se milte ho, isliye zyada crests per second tumhare ear se takraate hain → zyada $f^{\prime }$. Formula mein, source still matlab denominator plain $v$ hai; toward matlab upar $+v_o$: $f^{\prime }=f\frac{v+v_o}{v}>f.$

Observer still ⇒ top plain $v$ hai. Source away ⇒ wavelength stretch ho rahi hai, pitch kam ho rahi hai, isliye humein woh sign chahiye jo $f^{\prime }$ ko chhota kare: woh hai denominator mein $+v_s$. $f^{\prime }=f\frac{v}{v+v_s}.$ Logic check karo: bada denominator matlab chota $f^{\prime }$ = kam pitch, exactly wahi jo receding ko karna chahiye. ✓

Source still, observer toward ⇒ upar $+v_o$: $f^{\prime }=680\cdot \frac{340+34}{340}=680\cdot \frac{374}{340}=680\cdot 1.1=748\text{Hz}.$ $680$ se zyada ✓ (tum approach kar rahe ho). Wavefront picture dekho

Observer still ⇒ upar plain $v$. Source toward ⇒ wavelength squeeeze ho rahi hai ⇒ denominator mein $-v_s$: $f^{\prime }=660\cdot \frac{340}{340-30}=660\cdot \frac{340}{310}=660\cdot 1.09677\dots =723.9\text{Hz}.$ Figure

Source still ⇒ denominator plain $v$. Observer away ⇒ tum crests ke saath bhag rahe ho, isliye woh tumhe zyada dheere overtake karte hain ⇒ upar $-v_o$ (recession pitch ghata hai): $f^{\prime }=f\frac{v-v_o}{v}=500\cdot \frac{340-17}{340}=500\cdot \frac{323}{340}=500\cdot 0.95=475\text{Hz}.$ $500$ se kam ✓. Dhyan do yeh L2.1 ka exact mirror hai — wahi mechanism (crest-arrival speed), opposite sign.

Receding source wavelength stretch karta hai ⇒ denominator mein $+v_s$: $f^{\prime }=660\cdot \frac{340}{340+30}=660\cdot \frac{340}{370}=660\cdot 0.918918\dots =606.5\text{Hz}.$ Poora "pass-by" drop hai $723.9\to 606.5$ Hz — woh girta hua wail jo tum sunते ho jab woh guzrta hai.

Dono motions "toward" hain, dono pitch badhate hain ⇒ upar $+v_o$, neeche $-v_s$: $f^{\prime }=500\cdot \frac{340+10}{340-30}=500\cdot \frac{350}{310}=500\cdot 1.129032\dots =564.5\text{Hz}.$ Akele kisi ek se bada shift ✓.

Tum source ki taraf move kar rahe ho ⇒ upar $+v_o$. Source door ja raha hai ⇒ denominator mein $+v_s$. $f^{\prime }=700\cdot \frac{340+25}{340+40}=700\cdot \frac{365}{380}=700\cdot 0.960526\dots =672.4\text{Hz}.$ Thoda ghatti hai. Forecast check: tum peecha kar rahe ho lekin bhagte source ko pakad nahi pa rahe, isliye recession thoda jeet jaata hai — ek chota sa drop, exactly wahi jo mila ✓.

Hop 1 — wall ek moving observer hai jo still horn ke paas aa rahi hai, isliye uski observer speed $v_o=u$ hai: $f_1=f\frac{v+v_o}{v}=f\frac{v+u}{v}=500\cdot \frac{360}{340}.$ Hop 2 — wall ab ek moving source ki tarah kaam karti hai ($f_1$ emit karte hue) still horn ki taraf approach karti hai, isliye uski source speed $v_s=u$ hai: $f^{\prime }=f_1\frac{v}{v-v_s}=f_1\frac{v}{v-u}=500\cdot \frac{v+u}{v}\cdot \frac{v}{v-u}=500\cdot \frac{v+u}{v-u}.$ $f^{\prime }=500\cdot \frac{360}{320}=500\cdot 1.125=562.5\text{Hz}.$ Clean result dekho: still source ki taraf approach karne wale reflector ke liye, shift hai $\frac{v+u}{v-u}$ (dono effects stack hote hain, ek baar sunne ke liye, ek baar re-emitting ke liye). Yahan $u$ sirf wall ki ground speed hai, jo hop 1 mein $v_o$ aur hop 2 mein $v_s$ ka role play karti hai.

Wind har jagah effective wave speed shift karti hai: $v\to v+w$, jahan $w$ positive hai agar air observer ki taraf drift kare aur negative hai agar source ki taraf drift kare. Crest speeds add kyun hoti hain, yeh samajhne ke liye Relative velocity dekho. Source still ⇒ effective-$v$ upar; observer toward ⇒ $+v_o$: $f^{\prime }=f\frac{(v+w)+v_o}{v+w}.$ (a) Tailwind, $w=+10$: $f^{\prime }=400\cdot \frac{350+20}{350}=400\cdot \frac{370}{350}=422.9\text{Hz}.$ (b) Headwind, $w=-10$: $f^{\prime }=400\cdot \frac{330+20}{330}=400\cdot \frac{350}{330}=424.2\text{Hz}.$ (Compare karo no-wind $f^{\prime }=400\cdot 360/340=423.5$ Hz se. Tailwind top aur bottom dono inflate karta hai, fractional shift shrink karta hai; headwind dono deflate karta hai, ise enlarge karta hai. $w$ ka sign poori kahani hai — yeh kabhi $v_o$ ko touch nahi karta.)

Approaching horn (source toward ⇒ denominator mein $-v_s$): $f_1=340\cdot \frac{340}{340-2}=340\cdot \frac{340}{338}$. Receding horn (source away ⇒ $+v_s$): $f_2=340\cdot \frac{340}{340+2}=340\cdot \frac{340}{342}$. $f_1=342.0118\dots \text{Hz},f_2=337.9532\dots \text{Hz}.$ Beat frequency = $|f_1-f_2|$: $f_{\text{beat}}=342.0118-337.9532=4.06\text{Hz}.$ Toh lagbhag 4 beats per second — ek dheemi throbbing.

Observer still, source approaching: $f^{\prime }=f\frac{v}{v-v_s}\Rightarrow v-v_s=\frac{fv}{f^{\prime }}=\frac{500\cdot 340}{540}.$ $v-v_s=314.8148\dots \text{m/s}\Rightarrow v_s=340-314.8148=25.19\text{m/s}.$ Toh horn listener ki taraf lagbhag $25.2$ m/s par move kar raha hai.

Source approaching, observer still: $f^{\prime }=f\frac{v}{v-v_s}$.

• $v_s=300$ par: $f^{\prime }=1000\cdot \frac{340}{40}=8500\text{Hz}$.
• $v_s=339$ par: $f^{\prime }=1000\cdot \frac{340}{1}=340000\text{Hz}$. Jaise jaise $v_s\to v=340$, denominator $\to 0$ aur $f^{\prime }\to \infty$: har crest essentially usi jagah emit hota hai jahan pichla arrive hota hai, isliye saare wavefronts ek single front par pile ho jaate hain. Woh front shock wave / sonic boom hai — Doppler formula ab valid nahi hai kyunki source ke aage ab koi well-defined wavelength nahi hai. $v_s>v$ ke liye source apni khud ki sound se aage nikal jaata hai aur ordinary formula ek negative frequency deta hai (physically meaningless). Dekho Sonic boom and shock waves.

(a) $f^{\prime }=400\cdot \frac{340+40}{340}=400\cdot \frac{380}{340}=447.06\text{Hz}.$ (b) $f^{\prime }=400\cdot \frac{340}{340-40}=400\cdot \frac{340}{300}=453.33\text{Hz}.$ Kyun alag: observer case additive hai — yeh sirf crest arrival speed up karta hai, top line. Source case wavelength par multiplicative hai — yeh physically spacing shrink karta hai, bottom line. Algebraically $\frac{v+u}{v}$ vs $\frac{v}{v-u}$ sirf $u/v$ mein first order tak agree karte hain; source term mein extra "compounding" hai (shrinking har crest ke liye repeat hoti hai), isliye woh hamesha thoda zyada shift karta hai. Yeh asymmetry exactly wahi hai kyun acoustic Doppler effect (medium exist karta hai) symmetric nahi hai, unlike Doppler effect of light, jahan koi medium nahi hota aur sirf relative velocity matter karti hai.

Directly, $f^{\prime }=f\frac{v+0}{v-0}=f$ — koi motion nahi, koi shift nahi ✓. Small $u$ ke liye (dono toward): $f^{\prime }=f\frac{v+u}{v-u}$. $\frac{v+u}{v-u}\approx \left(1+\frac{u}{v}\right)\left(1+\frac{u}{v}\right)\approx 1+\frac{2u}{v}$ use karte hue, fractional shift hai $\frac{f^{\prime }-f}{f}\approx \frac{2u}{v}.$ Concretely $u=5$ m/s par: $\frac{2\cdot 5}{340}=0.0294=2.94\%$. Toh do objects jo combined $10$ m/s par close ho rahe hain, roughly $3\%$ pitch rise produce karte hain — clearly audible shift.

$\frac{v+u}{v-u}$ kyun hai Figure dekho: wall ek approaching horn sunti hai (moving observer, factor $\frac{v+u}{v}$), phir jo usne suna use re-emit karti hai jab woh khud horn ki taraf approach kar rahi hai (moving source, factor $\frac{v}{v-u}$). Dono hops multiply karo aur $v$ cancel ho jaate hain: $\frac{v+u}{v}\cdot \frac{v}{v-u}=\frac{v+u}{v-u}$.